The domain of a function is the set of values for which the function is evaluable. To guarantee that, we must make sure that:
- No fraction has a zero denominator
- No square root has a (strictly) negative argument
- No logarithm has a negative (or zero) argument.
So, in your case, you must make sure that the fraction #12/x# is defined, and that the square root is defined as well.
The first request is quite easy, since #12/x# is a non-zero-denominator fraction if and only if #x\ne 0#
As for the whole root, we must make sure that #12/x+9# is greater to zero, or at most exactly zero. In fact, if we would choose a value of #x# such that #12/x+9<0#, we would be calculating the square root of a negative number, which is impossible to do using real numbers.
Now, if #x# is positive, #12/x# is positive as well, and so #12/x+9# will be positive because it's the sum of two positive numbers.
If #x# is negative, we must solve #12/x+9 \ge 0#. Subtracting #9# from both sides, we get #12/x\ge -9#. Since #x\ne 0#, we can multiply both terms by #x#, but since #x# is negative we must invert the inequality, obtaining #12\le -9x#; and again dividing by #-9# both terms, we have #12/{-9}\ge x#, which means #x \le -4/3#.
So, every positive number is ok, and amongst the negative ones we can accept only those which are lesser or equal to #-4/3#. This means that the domain of the function is
#D={ x \in \mathbb{R}:\ x\le -4/3}\cup { x \in \mathbb{R}:\ x> 0} = (-infty, -4/3] \cup (0,\infty)#
